3)a/$201^{2}$ = $(200+1)^{2}$ = $200^{2}$+2.200+1 = 40000+400+1 = 40401
d/ $2016^{2}$-2015.2017 = ($2016^{2}$-1) -2015.2017+1 = (2016-1)(2016+1) -2015.2017 +1 =1
b/ $498^{2}$ = $(500-2)^{2}$ = $500^{2}$-2.500.2+$2^{2}$ = 250000-2000+4=248004
c/ 93.107 = (100-7)(100+7)= $100^{2}$ - $7^{2}$ = 10000 - 49 = 9951
e/ $\frac{2016^{3}-1}{2016^{2}+2017}$ = $\frac{(2016-1)(2016^{2}+2016+1)}{2016^{2}+2017}$ =2015
h/M-N=$99^{2}$ - $98^{2}$ + $97^{2}$ ...-$2^{2}$ +1
= ($99^{2}$-$98^{2}$)- ($97^{2}$-$96^{2}$)...+($3^{2}$ - $2^{2}$)+1
=(99+98)(99-98)+(97+96)(97-96)...(3+2)(3-2)+1
=99+98+97+96...+3+2+1
=(99+1)(99/2)=4950
M-N+50=5000
g/($2016^{2}$ - $2015^{2}$) + ($2014^{2}$ -$2013^{2}$)+($2012^{2}$-$2011^{2}$)+($2^{2}$-1)
=(2016+2015)(2016-2015)+(2014+2013)(2014-2013)+(2012+2011)(2012-2011)+(2+1)(2-1)
=2016+2015+2014+2013+2012+2011+2+1=12084
4)a/$x^{2}$ -x+1=$x^{2}$ -2.$\frac{1}{2}$ .x +$(\frac{1}{2})^{2}$ + $\frac{3}{4}$ =$(x-\frac{1}{2})^{2}$+$\frac{3}{4}$
b/$x^{2}$ +3x+4=$x^{2}$ +2.$\frac{3}{2}$ .x +$(\frac{3}{2})^{2}$ + $\frac{7}{4}$ =$(x+\frac{3}{2})^{2}$+$\frac{7}{4}$
c/$x^{4}$ -4$x^{2}$+6=$(x^{2})^{2}$ -2.2$x^{2}$ +$2^{2}$ +2 =$(x^{2}-2)^{2}$+2
d/2$x^{2}$ +6x-15=2($x^{2}$ +2.$\frac{3}{2}$ .x +$(\frac{3}{2})^{2}$) - $\frac{39}{2}$ =2$(x+\frac{3}{2})^{2}$-$\frac{39}{2}$
e/$x^{2}$ -5x+9=$x^{2}$ -2.$\frac{5}{2}$ .x +$(\frac{5}{2})^{2}$ + $\frac{11}{4}$ =$(x-\frac{5}{2})^{2}$+$\frac{11}{4}$
g/3$x^{2}$ -12x+20=3($x^{2}$ -2.2.x +$2^{2}$) + 8 =3$(x+2)^{2}$+8
5)a/$x^{2}$+10x+26+$y^{2}$+2y=( $x^{2}$+10x+25)+($y^{2}$+2y+1)=$(x+5)^{2}$+$(y+1)^{2}$
b/$a^{2}$-6a+5-$b^{2}$-4b=( $a^{2}$-6a+9)-($b^{2}$+4b+4)=$(a-3)^{2}$-$(b+2)^{2}$
c/$x^{2}$-2xy+2$y^{2}$+2y+1=( $x^{2}$-2xy+$y^{2}$)+($y^{2}$+2y+1)=$(x-y)^{2}$+$(y+1)^{2}$
d/4$x^{2}$-12x-$y^{2}$+2y+8=[$(2x)^{2}$-2.3.2x+9)]-($y^{2}$-2y+1)=$(2x-3)^{2}$-$(y+1)^{2}$
